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Direction of propagation of an EM wave

  1. Aug 5, 2012 #1
    I'm reading my course book on ELectromagnetism and it is talking about a wave moving in the y-z plane but with polarisation in the x-direction, and it says that the equation

    $$\mathbf{E}=E_0 2i sin(k_0 z\ cos \theta) exp[i(k_0 y\ sin \theta - \omega t)]\mathbf{e}_x$$

    Shows that there is no propagation in the z-direction, but that instead a standing wave is formed when he wave bounces at an oblique angle between the two perfect conductors at z=0 and z=-a.

    My question is...if the wave equation contains a factor involving z, how is it not propagating in the z direction? I suspect there is something I haven't understood about wave propagation and movement.

    Thanks in advance
     
  2. jcsd
  3. Aug 5, 2012 #2

    gabbagabbahey

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    What variable(s) must change as you progress forward (or backward) in time, If the amplitude of the electric field remains unchanged? Do you see how this describes a specific point on the wavepulse travelling through space as time progresses?
     
  4. Aug 6, 2012 #3
    Ah, well I can see that the term in the exponential brackets must be constant in order for there to be no attenuation over time, so I imagine that y must increase as t increases...hence propagation in the y-direction. So far so good.
    So I guess at z=0 the expression is zero, and this indicates that the wave has a node at this point? This might sound like a stupid question, but what happens if the wave reaches the boundary of the conductor at a point where it doesn't have a non-zero value? Is it absorbed by the conductor?

    Thanks again for your help
     
  5. Aug 6, 2012 #4

    gabbagabbahey

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    If you are only looking at constant [itex]z[/itex], or [itex]\theta = \frac{\pi}{2}[/itex], then yes. In general however, all you can say is that [itex]\sin(k_0 z \cos \theta ) e^{i ( k_0 y \sin \theta - \omega t )}[/itex] must remain constant. There can certainly be propagation in the z-direction for this waveform.

    I assume there must be more to the situation discussed in your text than what you've provided. Which text is this from?
     
  6. Aug 7, 2012 #5

    Born2bwire

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    I would put it down as being that there are two possible solutions to the wave equation here whereby the standing wave is the superposition of these solutions. As you stated previously, the standing wave is the superposition of a wave bouncing back and forth. So in the context of your overall wave solution, what you have is, for example, a wave that is moving in the y and z direction at some point Z and a wave that is moving in the y and -z direction at the same point Z. The superposition of these two gives rise to the standing wave in the z direction, but since they are both solutions, you can think of them as being separate waves if you like. If you were to excite the actual mode between the two plates, you could do so in such a way so that there is only one of these solutions excited at a time.

    Of course, the underlying reason why there is no propagation in the z direction (more accurately net propagation) is that the wave is confined in the z direction due to the boundary conditions of the plates. So the fact that the full solution exhibits a standing wave is just demonstrating how the net propagation of the wave is guided in the y direction and confined in the z direction by the plates.
     
  7. Aug 7, 2012 #6
    @Gabba:

    I'm not sure if I've missed out any relevant details. The text is a university course book. I won't say which university, but it is a respectable publication. It has a section on normal incidence of EM waves on conductors, then a shorter section on oblique incidence, which this is taken from. It derives the result I gave in my original post, and then states that:
    "In contrast to section...where we found a stationary wave for normal incidence, in the case of oblique incidence, there is now a propagating wave. In fact, the exponential term shows that the wave propagates in the y-direction, parallel to the conducting boundary, with speed omega/(k_0 sin theta)."
    It goes on to talk about modes where propagation can satisfy the boundary conditions for the parallel component of the electric field.
    You say the wave does propagate in the z-direction because of the z-term. Doesn't this just affect the amplitude of the wave?

    @Born2

    It sounds like you're right, as the next section is about waveguides following this initial model and applying it to physical constraints in the x-y and x-z planes. But as I visualise this as a wave bouncing off the walls between z=0 and z=-a, is it not propagating in the z-direction? (i.e. if there were no constraints, would it move in the z-direction?

    I'm confused because the equation seems to suggest movement (propagation?) in the y-direction, with amplitude which depends on the z-variable. But if the wave incides on the conducting boundary at z=0 and z=-a then it must have moved in the z-direction, musn't it?

    Apologies if this sounds garbled.
     
  8. Aug 7, 2012 #7

    Born2bwire

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    It is propagating in the z direction, but this propagation isn't a net propagation because it is just bouncing back and forth between z=0 and z=-a. Take a look at slide 135 in this set:

    http://www.amanogawa.com/archive/docs/EM12.pdf

    This slide shows the actual path of propagation for the plane wave inside a parallel plate waveguide. As you suspect, the wave bounces back and forth between the two plates. The catch here is that the path that they dictate is just one of TWO equally valid solutions. The other solution has the wave going along the same angles, but it starts out bouncing off of the other plate. When you take the superposition of the two solutions, the result looks like a standing wave in the z-direction.

    When you solve for the modes of the parallel plate waveguide, you are solving for all possible fields. So the solution represents a superposition of all the possible waves. When they talk about the direction of propagation for a waveguide, notation often gets lazy. The whole purpose of a waveguide is to direct the propagation of energy. As such, people are focused on the direction of guided propagation, not necessarily that actual path of propagation. As such, for a parallel plate wave guide then the direction of guided or net propagation in your case is along the y direction. There is no net flow of energy in the z direction due to the confinement by the plates. Also note that this looseness of terms may give rise to some confusion when they talk about the waveguide modes. Often you will hear, like in the case of a rectangular waveguide, that only TE or TM modes exist, not TEM. This seems to go against our knowledge of Maxwell's Equations which states that EM waves are TEM. What they really mean, is that there are no TEM modes with respect to the guided direction of propagation. So they are not talking about TE, TM or TEM regarding the instantaneous direction of propagation but the guided direction of propagation.

    Without the confinement of the plates, then you could of course get propagation in the z-direction. Without the plates the solution is just the trivial free-space solution which is a superposition of plane waves in any direction.

    Also note that your solution is the time-harmonic solution. It is not the time-domain solution so it is not going to represent a causal propagation of a wave. To get the actual time-domain solution, you have to take the Fourier transform or the real part of the time-harmonic solution (depending upon how you define the time-harmonic case). Despite this, a basic knowledge of the time-harmonic plane wave allows us to easily infer how the plane wave would propagate in this case based off of the time-harmonic solution.
     
  9. Aug 7, 2012 #8

    gabbagabbahey

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    Okay, as Born2 says, if there are two perfect conducting planes at z=0 and z=-a, then the given waveform is only the electric field between those two planes, outside it will be zero (barring additional sources there). This means that propagation in the z-direction will average out (as the waveform's motion in that direction will cause it to just reflect back and forth between the planes) - there will be a standing wave there.

    However, this is not evident in the waveform you give. It is only after you apply the boundary conditions on [itex]z[/itex] that the result shows explicitly this behaviour.

    Look at the real part of [itex]\mathbf{E}[/itex] and remember that [itex]\sin u \cos u = \frac{1}{2} \left[ cos(u-v) + \cos(u+v) \right][/itex]
     
  10. Aug 8, 2012 #9
    Thanks very much for the clear and insightful replies. I'm not familiar with time-harmonic vs time-domain solutions but I will investigate, rather than ask someone to explain a whole new concept from scratch.
    I think I have understood the basics now.
    Thanks again.
     
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